I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).

Let P be the proposition that the sum of the first n even numbers is n(n+1); P(n) : 2 + 4 + 6+ … + 2n = n(n+1)

With mathematical induction we first need to show that this is true when n=1Since we are only discussing even numbers, we will start with 2:2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1: 2=1(2) and 2=2

(Here I was told I was wrong again, and it was suggested that I add 2(k+1) to both sides, but I don't see how that works out..... help I'm stuck!)

THanks.

Oct 30th 2008, 10:43 AM

Chris L T521

Quote:

Originally Posted by Conorsmom

Hi there,

I have this assignment, and I thought I was on the right path, but when I received it back with suggestions to fix it, I'm still not sure where to go from here. Thanks for your help!

Create a proof by mathematical induction that demonstrates that the sum of the first n even numbers is equal to n(n + 1).

Let P be the proposition that the sum of the first n even numbers is n(n+1); P(n) : 2 + 4 + 6+ … + 2n = n(n+1)

With mathematical induction we first need to show that this is true when n=1Since we are only discussing even numbers, we will start with 2:2=n(n+1); n=1 so; 2=1(1+1), following simple algebra we find that this is true for n=1: 2=1(2) and 2=2